What is the inverse of (x^2-9)/(7x^2) ?

The inverse of (x^2-9)/(7x^2) is sqrt(-9/(7x-1)),-sqrt(-9/(7x-1))

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inverse of (x^2-9)/(7x^2)

inverse

\frac{x ^{2} - 9}{7 x ^{2}}

- \frac{9}{7 x - 1}, - - \frac{9}{7 x - 1}

Solution steps

y = \frac{x ^{2} - 9}{7 x ^{2}}

Replace

x

with

y

x = \frac{y ^{2} - 9}{7 y ^{2}}

Solve for

y, x = \frac{y ^{2} - 9}{7 y ^{2}}

- \frac{9}{7 x - 1}, - - \frac{9}{7 x - 1}

f (x) = \frac{( - 3 - 4 x )}{2 + 2 x}

lo g_{10} (x + 3) - 3

lo g_{10} (x + 3) - 2

f (x) = 7^{x} + 19

f (x) = 3 x + 9 + 2

What is the inverse of (x^2-9)/(7x^2) ?
The inverse of (x^2-9)/(7x^2) is sqrt(-9/(7x-1)),-sqrt(-9/(7x-1))